Student Handout 13 2014
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Transcript of Student Handout 13 2014
Differential equations of motion
CHEE 3363Spring 2014Handout 13
�Reading: Fox 5.1
Brief reminder: Taylor series expansion:
f(x0 + ∆x) = f(x0) +df
dx
∣
∣
∣
∣
x=x0
∆x + . . .
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Learning objectives for lecture
1. State conservation of mass in differential form.�
2. Find velocity components that satisfy conservation of mass.
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Recall: Integral formulation of lawsMass balance:
Linear momentum balance:
Angular momentum balance:
Energy balance:
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∂
∂t
∫
CV
ρdV = −
∫
CS
ρv·dA
F = FS + FB =∂
∂t
∫
CV
vρdV +
∫
CS
vρv·dA
r × Fs +
∫
M(sys)
r × g dm + Tshaft =∂
∂t
∫
CV
r × vρ dV +
∫
Cs
r × vρv·dA
Q − Ws − Wshear − Wother =∂
∂t
∫
CV
eρ dV +
∫
CS
(u + pv +v2
2+ gz)ρv · dA
dx
dy
dz
Conservation of mass: differential form 1
x
y
z
Want: to obtain a differential version of the
Density at center: ρ
Velocity at center: (u,v,wu
v
w∂
∂t
∫
CV
ρdV = −
∫
CS
ρv·dA
First stepρ, (u,v,w
keep 1st order only:
Remember: all functions evaluated at the center point!�4
ρ|x+dx/2
= ρ +
(
∂ρ
∂x
)
dx
2+
(
∂2ρ
∂x
)
1
2!
(
dx
2
)2
+ . . .
u|x+dx/2
= u +
(
∂u
∂x
)
dx
2
Conservation of mass: differential form 2
ρ, (u,v,wkeep 1st order only:
∫
CS
ρv · dAEvaluate
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[
ρ +
(
∂ρ
∂x
)
dx
2
] [
u +
(
∂u
∂x
)
dx
2
]
dydz
= ρu dy dz +1
2
[
u
(
∂ρ
∂x
)
+ ρ
(
∂u
∂x
)]
dx dy dz
u|x−dx/2
= u +
(
∂u
∂x
) (
−
dx
2
)
ρ|x−dx/2
= ρ +
(
∂ρ
∂x
) (
−
dx
2
)
Conservation of mass: differential form 3
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= −ρu dy dz +1
2
[
u
(
∂ρ
∂x
)
+ ρ
(
∂u
∂x
)]
dx dy dz
−
[
ρ −
(
∂ρ
∂x
)
dx
2
] [
u −
(
∂u
∂x
)
dx
2
]
dydz
from area normal
Conservation of mass: differential form 4Exercise for the reader: write out equations for other four faces
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∫
CS
ρv · dA =
[(
u
(
∂ρ
∂x
)
+ ρ
(
∂u
∂x
))
+
(
v
(
∂ρ
∂y
)
+ ρ
(
∂v
∂y
))
+
(
w
(
∂ρ
∂z
)
+ ρ
(
∂w
∂z
))]
dx dy dz
∫
CS
ρv · dA =
[
∂ρu
∂x+
∂ρu
∂y+
∂ρu
∂z
]
dx dy dz
Conservation of mass: differential form 5
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Putting it all together in rectilinear coordinates:
∂
∂t
∫
CV
ρ dV =∂ρ
∂tdx dy dz
∂ρu
∂x+
∂ρu
∂y+
∂ρu
∂z+
∂ρ
∂t= 0
@⇢u
@x
+@⇢v
@y
+@⇢w
@z
=
Conservation of mass: differential form 6We can rewrite the conservation laws using the vector operators:
and put this all together to obtain:
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∇·ρv +∂ρ
∂t= 0
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For an (ρ
For (ρ = ρ(
∇ · v = 0
∇·ρv = 0
∇·ρv +∂ρ
∂t= 0
Vector calculus derivationNote: using vector calculus theorems, we can derive this much faster:
Rewrite 2nd term slightly:
Recall Gauss’s divergence theorem:
and apply to mass balance equation:
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∂
∂t
∫
CV
ρ dV +
∫
CS
ρv · dA = 0
∫
CS
ρv · dA =
∫
CS
ρv · ndA
∫
surf
F · ndA =
∫
vol
∇ · FdV
∂
∂t
∫
CV
ρ dV +
∫
CS
ρv · dA =
∫
CV
∂ρ
∂tdV +
∫
CV
∇ · vdV = 0
1
Wanted: simplest form of the y component of velocity satisfying continuity?
Assumptions:
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u = A exp⇣x
b
⌘cos
⇣y
b
⌘
2
Wanted: simplest form of the y component of velocity satisfying continuity?
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u = A exp⇣x
b
⌘cos
⇣y
b
⌘
Integrate to get v(x,
Conserv. of mass in cylindrical coord.s 1Consider a piece of a cylindrical differential control volume as shown.��Want:through this surface.
∫
CS
ρv · dA
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Inside r = −
[
ρ −
(
∂ρ
∂r
)
dr
2
] [
vr −
(
∂vr
∂r
)
dr
2
] (
r −dr
2
)
dθ dz
Outside (r =
[
ρ +
(
∂ρ
∂r
)
dr
2
] [
vr +
(
∂vr
∂r
)
dr
2
] (
r +dr
2
)
dθ dz
Conserv. of mass in cylindrical coord.s 2Consider a piece of a cylindrical differential control volume as shown.��Want:through this surface.
∫
CS
ρv · dA
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Back (θ =
[
ρ +
(
∂ρ
∂θ
)
dθ
2
] [
vθ +
(
∂vθ
∂θ
)
dθ
2
]
dr dz
Front (-θ = −
[
ρ −
(
∂ρ
∂θ
)
dθ
2
] [
vθ −
(
∂vθ
∂θ
)
dθ
2
]
dr dz
Conserv. of mass in cylindrical coord.s 3Consider a piece of a cylindrical differential control volume as shown.��Want:through this surface.
∫
CS
ρv · dA
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Bottom ( = −
[
ρ −
(
∂ρ
∂z
)
dz
2
] [
vz −
(
∂vz
∂z
)
dz
2
]
rdθ dr
Top ( =
[
ρ +
(
∂ρ
∂z
)
dz
2
] [
vz +
(
∂vz
∂z
)
dz
2
]
rdθ dr
Conserv. of mass in cylindrical coord.s 4
Recall: the form of the vector operator del in cylindrical coordinates:
v = ervr + eθvθ + kvz
Result:
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∇ = er
∂
∂r+ eθ
1
r
∂
∂θ+ k
∂
∂z
1
r
∂(rρvr)
∂r+
1
r
∂(ρvθ)
∂θ+
∂(ρvz)
∂z+
∂ρ
∂t= 0
∂er
∂θ= eθ
∂eθ
∂θ= −erand
Example: cylindrical coordinates 1
Wanted: simplest form of the θ component of velocity satisfying continuity
Assumptions:
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vr = U cos θ
(
1 −
a2
r2
)
vr = U cos θ
(
1 −
a2
r2
)
Example: cylindrical coordinates 2
Wanted: simplest form of the θ component of velocity satisfying continuity
Integrate to get vθ(r,θ :
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